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Dilution and Concentration - Lippincott Williams & …

Chapter 10207 Dilution and ConcentrationOBJECTIVESUpon completion of this chapter, the technician student willbe able to: Describe the relationship of active ingredients and diluents if the amount of active ingredient remains constant and the amount of diluent is increased ordecreased. Determine the percent strength and ratio strength of a given product when the active ingredient remains constant and the amount of diluent is increased ordecreased. Determine the volume of solution of a desired strengthgiven a specified quantity of any given strength. Determine the volume of a specified stock solutionneeded to prepare a given solution.

Solve by Inverse Proportion: Calculate the Active Ingredient: 50 mL of a 1:20 solution contains 2.5 g of aluminum acetate. 25 1 1000 400.g g mL mL mL Ratio strength

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Transcription of Dilution and Concentration - Lippincott Williams & …

1 Chapter 10207 Dilution and ConcentrationOBJECTIVESUpon completion of this chapter, the technician student willbe able to: Describe the relationship of active ingredients and diluents if the amount of active ingredient remains constant and the amount of diluent is increased ordecreased. Determine the percent strength and ratio strength of a given product when the active ingredient remains constant and the amount of diluent is increased ordecreased. Determine the volume of solution of a desired strengthgiven a specified quantity of any given strength. Determine the volume of a specified stock solutionneeded to prepare a given solution.

2 Determine the quantity of an active ingredient in a specified amount of solution needed to prepare a given solution. Define the alligation methods of problem solving. Use the alligation methods (alligation alternate and alligation medial) to determine the percent strength of alcohol mixtures. Use the alligation methods (alligation alternate and alligation medial) to determine relative amounts of components mixed together to make a mixture of arequired Alligation Alligation alternate Alligationmedial Diluent Stock solution456X789 % 11/20/06 2:45 PM Page 207 Chapter 9 addresses problems arising from the quantitative relationship betweenspecific ingredients and the pharmaceutical preparation as a whole.

3 This chapterintroduces problems of a slightly different character: those that arise when pharma-ceutical preparations are diluted(by the addition of diluentor by admixture withsolutions or mixtures of lower strength) or are concentrated(by the addition ofactive ingredient or by admixture with solutions or mixtures of greater strength).Problems such as these sometimes seem complicated and difficult. Solving someof these calculations requires a series of steps. Understanding the basic concepts ofdilution and Concentration uncomplicates these problems can be solved in several ways.

4 The best way is not necessarilythe shortest: the best way is the one that is clearly understoodand that leads to thecorrect substance addedto a pharmaceutical productto reduce its strength, ordilute it. A diluent most oftenhas no drug substance in it;examples are sterile waterand Calculations for the Pharmacy Technician208456X123 0 =+789 % to Simplify the CalculationThese two rules, wherever they may be applied, greatly simplify the ratio strengths are given, convert them to percentage strengths before settingup a is much easier to solve using a decimal or percent, like 10 (%) : (%)than a fraction represented by the ratio like 1 10:1 proportional parts enter into a calculation, reduce them to lowest of calculating with a ratio like 25 (parts) : 75 (parts),simplify it to 1 (part) : 3 (parts).

5 Pharmaceutical preparationbefore dilutionPharmaceutical preparationafter Dilution (same amountof active drug although the volume has doubled)Relationship Between Strength and Total QuantityIf a mixture of a given percentage or ratio strength is diluted to twice its originalquantity, its active ingredient will be contained in twice as many parts of the whole,and its strength therefore will be reduced by one-half. So if 50 mL of a solution con-taining 10 g of active ingredient with a strength of 20% or 1 : 5 w/v is diluted to 100 mL, 11/20/06 2:45 PM Page 208the original volume is doubled, but the original strength is now reduced by one-halfto 10% or 1:10 , then, the amount of active ingredient remains constant, any change in thequantity of a solution or mixture of solids is inversely proportional to the percent-age or ratio strength;that is, the percentage or ratio strength decreases as the quan-tity increases, and vice versa.

6 In other words, if the amount of active ingredientremains the same and the volume gets larger, the Concentration gets smaller. Like-wise, if the amount of active ingredient remains the same and the volume getssmaller, the Concentration relationship is generally true except for volume-in-volume and weight-in-volume solutions containing components that contract when mixed together. It is notpossible to add the volume of ingredients and get the total volume of the final prod-uct in all cases. It is possible when mixing in this chapter generally may be solved by one of the following methods:1.

7 Inverse proportion:where C = Concentration and Q = The following formula:That is,3. Determining the quantity of active constituent (solute) needed and then calcu-lating the quantity of the available solution (usually concentrated or stock solu-tion) that will provide the needed amount of most situations the student technician is encouraged to use the formulamethod of solving these Dilution and Concentration problems. Sometimes the thirdmethod is the easiest of the three, usually when the strength of the product is verysmall and the technician is using a prepared strength available in a vial or of these are detailed later in the chapter.

8 In the formula method the sec-ond quantity and second Concentration are always that of the final product, not of theactive ingredient or and Concentration of LiquidsDetermination of Percentage or Ratio StrengthCalculating the percentage or ratio strength of a solution made by diluting or con-centrating (by evaporation) a solution of given quantity and strength entails C11 2 2 (()()quantityquantityconcentration11 222)() concentrationCC2QQ121 CHAPTER 10 Dilution and 11/20/06 2:45 PM Page 209 Solve by Inverse Proportion:Calculate the Active Ingredient:500 mL of 15% v/v solution contains 75 mL of methyl salicylate (active ingredient).

9 If 50 mL of a 1:20 w/v solution of aluminum acetate is diluted to 1000 mL, what isthe ratio strength w/v?50510000 251 400mLmL() ()=() ()==%%.% : ,XannswerQCQ C112 2 120 5:%=1500751005 mLmL==%%%,XXanswer1500500155 mLmL==%%%,XXanswer5001515005mLmL() ()=() ()=%%%,XXanswerQCQ C1122 Pharmaceutical Calculations for the Pharmacy Technician210 Water15%500 mL500 mLOriginal 15% concentration1500 mLAdded volume (from water)Examples:If 500 mL of a 15% v/v solution of methyl salicylate in alcohol is diluted to 1500 mL,what is the percentage strength v/v? 11/20/06 2:45 PM Page 210 Solve by Inverse Proportion:Calculate the Active Ingredient:50 mL of a 1:20 solution contains g of aluminum strength===XX11 400:,answer1000505100012050025mLmLmLmL== =%%.

10 %XXX====1 4001 4001 400:,:,answerXanswerCHAPTER 10 Dilution and Concentration211 Critical Thinking of a solid or liquid does notentail adding moredrug. Therefore, the amount of drug in the diluted productis the same;only the volume is a cup of coffee is too strong for a person s taste, theperson may add water to diluteit. No additional coffee is 0 =+789 % strong coffeeStrong coffeeBecause no additional coffee was added,the drink is now less Amount of Solution of a Desired StrengthCalculating the amount of solution of a desired strength that can be made by dilutingor concentrating (by evaporation) a specified quantity of a solution of given strengthinvolves the 11/20/06 2:45 PM Page 211 Examples:How much 10% w/w (in grams) ammonia solution can be made from 1800 g of 28% w/w strong ammonia solution?


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